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kodGreya [7K]
2 years ago
8

Which of the following functions is a direct variation?

Mathematics
2 answers:
Nezavi [6.7K]2 years ago
7 0

Answer:

f(x)=2x

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form y/x=k or y=kx

In a direct variation the constant of proportionality k is equal to the slope of the line m and the line passes through the origin

therefore

in this problem

f(x)=2x -----> represent a direct variation

valentina_108 [34]2 years ago
4 0
When two variable quantities have a constant (unchanged) ratio, their relationship is called a direct variation. It is said that one variable "varies<span> directly" as the other...simply when the y-intercept is zero.

So the best answer is f(x)=2x</span>
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What is 3 - 1 x 4 pls answer need help with this I’m getting -1 and 8
MatroZZZ [7]

Answer: -1

Step-by-step explanation:

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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

\ln(x) = \ln 3 + \ln(1+y).

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
3 years ago
Can someone help me solve these two problems? i need to find what x is.
Bond [772]

Answer:

1. x=2

2. x= 0.33

Step-by-step explanation:

1.So to start off you have a coefficient for 2 so you have to find what 2 to the power of 2 is =4 so then you have your new problem 4x+5=13 next you need to get 4x by itself so

-5 -5

then you have 4x=8. you divide by 4 on both sides

4. 4

so then you get x=2

2.So same here find what 3 to the power of 2 is =9

New equation 4x-1=9x+4x-4. Then combine like terms on the right you cant do it on the left because that is a separate part and those dont merge like terms so

9x+4x = 13x

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next subtract 4x on both sides

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next add 4 to both sides

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new equation. 3=9x

so then divide 9 on both sides

3÷9= 0.33

x= 0.33

Hope this made sense and helps

7 0
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